On this page
Haloalkanes are the first functional group where you have to decide which mechanism a reaction follows, and many questions are built around that decision. Students who memorise "tertiary means SN1" get the easy questions right and the rest wrong. This post gives you a four-question checklist, works through rate-law and stereochemistry examples, and covers the competition with elimination.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "nature of C-X bond" and "mechanisms of substitution reactions" in their unit on organic compounds containing halogens. The JEE (Advanced) 2026 brochure lists nucleophilic substitution reactions of alkyl halides "and their stereochemical aspects", plus rearrangements of alkyl carbocations. The NCERT Class 12 chapter on haloalkanes and haloarenes is the base text. Check the current year's syllabus.
What actually happens in SN2 and SN1?
SN2 (substitution, nucleophilic, bimolecular). NCERT describes the reaction of CH₃Cl with hydroxide as second order: the rate depends on both reactants. The nucleophile attacks the carbon from the side opposite the halogen, the new bond forms as the old one breaks, and there is no intermediate. The other three groups flip over like an umbrella in the wind, so the configuration is inverted.
Rate = k[RX][Nu⁻]
SN1 (substitution, nucleophilic, unimolecular). NCERT's example is tert-butyl bromide with hydroxide, which is first order. In the slow first step the C–Br bond breaks to give a carbocation; in the fast second step the nucleophile attacks it. Because the slow step involves only the haloalkane, the nucleophile's concentration does not appear in the rate law.
Rate = k[RX]
The planar carbocation can be attacked from either face, so an optically active haloalkane gives a racemic mixture.
The four-question checklist
What is the carbon? Methyl and primary halides go SN2, because there is room for the nucleophile to approach. Tertiary halides go SN1, because the carbocation is stable and the back side is blocked. Secondary halides can go either way. Allylic and benzylic halides form resonance-stabilised carbocations, so they react readily by SN1.
What is the nucleophile? A strong, usually negatively charged nucleophile (OH⁻, CN⁻, RO⁻, I⁻) at high concentration pushes towards SN2. A weak, neutral one such as water or an alcohol pushes towards SN1.
What is the solvent? NCERT notes that SN1 reactions are usually run in polar protic solvents such as water, alcohols or acetic acid, which stabilise the carbocation and the leaving halide ion. Polar aprotic solvents such as acetone favour SN2.
What is the leaving group? For both mechanisms, iodide leaves most easily and fluoride least: R–I > R–Br > R–Cl > R–F.
Worked example 1: reading the rate law
In an experiment, a haloalkane reacts with hydroxide. With both at 0.10 mol/L, the initial rate is 2.0 × 10⁻⁴ mol L⁻¹ s⁻¹. When [OH⁻] is doubled to 0.20 mol/L, the rate becomes 4.0 × 10⁻⁴ mol L⁻¹ s⁻¹. Which mechanism?
Doubling [OH⁻] doubled the rate, so the reaction is first order in hydroxide.
Hydroxide is in the rate law, so it takes part in the slow step: SN2.
k = rate / ([RX][OH⁻]) = 2.0 × 10⁻⁴ / (0.10 × 0.10) = 2.0 × 10⁻² L mol⁻¹ s⁻¹.
If the rate had stayed at 2.0 × 10⁻⁴ when [OH⁻] doubled, the mechanism would be SN1 and the substrate is likely tertiary.
Worked example 2: ranking reactivity
NCERT asks you to rank 1-bromobutane, 1-bromo-2-methylpropane, 2-bromobutane and 2-bromo-2-methylpropane. The answer it gives:
SN1: CH₃CH₂CH₂CH₂Br < (CH₃)₂CHCH₂Br < CH₃CH₂CH(Br)CH₃ < (CH₃)₃CBr
SN2: the exact reverse.
The reasoning: SN1 rate follows carbocation stability, tertiary over secondary over primary. SN2 rate follows crowding at the carbon; (CH₃)₂CHCH₂Br is primary, but the branch next door slows the backside attack compared with 1-bromobutane.
Worked example 3: stereochemistry
NCERT reports that (−)-2-bromooctane reacts with sodium hydroxide to give (+)-octan-2-ol, with the –OH on the side opposite where the bromine was. Since the reaction inverts the configuration at that carbon, it is SN2. If the same substrate were heated in water instead, the slow ionisation would give a carbocation and a racemic product.
The common mistake: assuming that a change in the sign of rotation, from (−) to (+), proves inversion. The sign of rotation and the arrangement of groups are separate properties. Our guide to stereoisomerism and chirality explains the difference.
When does elimination win?
Hydroxide and alkoxides are strong bases as well as nucleophiles. NCERT's summary: a primary halide prefers SN2; a secondary halide gives SN2 or elimination depending on the strength of the base; a tertiary halide gives SN1 or elimination. A bulky base prefers to pull off a proton rather than reach a crowded carbon. In practice, aqueous KOH tends to give substitution, and alcoholic KOH with heat tends to give elimination.
Why do students find this hard?
Cruz-Ramírez de Arellano and Towns (2014) interviewed undergraduates as they talked through alkyl halide reactions. Their gaps clustered in three places: classifying species as bases or nucleophiles, judging how strong a base or nucleophile is, and describing the steps and intermediates of the mechanism. The checklist above targets exactly those three. For more on working through mechanisms, see our post on learning organic mechanisms instead of lists.
If carbocation stability or nucleophile strength is the shaky part, the Fermi Concept Library covers every concept in chemistry, each with its own lesson, so you can go back to that idea on its own.
A practice routine for SN1 and SN2
Write the checklist on a card and use it for every question for a week.
Do ten "predict the mechanism and product" questions with varied substrates, nucleophiles and solvents.
Do four rate-law questions like worked example 1.
Do four stereochemistry questions, drawing the product's three-dimensional structure.
Mix in five substitution-or-elimination questions.
Today, apply the four questions to tert-butyl bromide in water and to bromomethane with sodium hydroxide, and write the rate law for each.
